Evaluating $\sum_{k=1}^{30} k(30-k)$

Question

I tried to rewrite it as $\sum_{k=1}^{30} k(30-\sum_{k=1}^{30}k)$ and then replace the $\sum_{k=1}^{30} k$ with $\frac{n(n+1)}{2}$ then substitute $n=30$ into the equation, however I am not getting the right answer.

Any help on how to solve this would be much appreciated.

Answer 1

You can't rearrange the sum like that. Instead you should write it as

$$\sum_{k=1}^{30}(30k-k^2) = \sum_{k=1}^{30}30k-\sum_{k=1}^{30}k^2.$$

From here, employ the expressions given by Sami.

As an aside, note that your summand is very symmetric. You can rewrite it as

$$\sum_{k=1}^{14}2k(30-k)+15^2.$$

Answer 2

Hint

These two equalities are useful:

$$\sum_{k=1}^nk=\frac{n(n+1)}2$$ and $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$$

Answer 3

For the sake of symmetry, let us sum from $k=0$ to $k=29$. By completing the square, we can see that we want $$\sum_1^{29}\left(15^2-(15-k)^2\right).$$ This is equal to $$(29)(15^2) -2\sum_{j=1}^{14} j^2.$$